Difference between revisions of "Radiators Temperature"
		
		
		
		
		
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| Line 11: | Line 11: | ||
Pulse width: ∆t= 50 ns=5*10-8 seconds  | Pulse width: ∆t= 50 ns=5*10-8 seconds  | ||
| − | So how many electrons we have in each second?  | + | So, how many electrons we have in each second?  | 
By Q=It, we have  | By Q=It, we have  | ||
| − | + |                                              N*e=f*I*∆t  | |
Where Ne is the total electron numbers hits target per second, e is electron charge and f, I and ∆t are given above.  | Where Ne is the total electron numbers hits target per second, e is electron charge and f, I and ∆t are given above.  | ||
So  | So  | ||
| − | N= f*I*∆t/e=1000*0.01*5*10-8/(1.6*10-19)=3.12075*1012  | + |                       N= f*I*∆t/e=1000*0.01*5*10-8/(1.6*10-19)=3.12075*1012  | 
So, we have around 3.12075*1012 electrons hit radiator per second.  | So, we have around 3.12075*1012 electrons hit radiator per second.  | ||
Revision as of 14:33, 29 May 2008
Calculation of Equilibrium temperature of Radiators
1.Calculating number of particles per second
We have electron beam of:
Frequency: f=1000Hz
Peak current: I=10mAmp=0.01 Amp
Pulse width: ∆t= 50 ns=5*10-8 seconds
So, how many electrons we have in each second?
By Q=It, we have
N*e=f*I*∆t
Where Ne is the total electron numbers hits target per second, e is electron charge and f, I and ∆t are given above. So
N= f*I*∆t/e=1000*0.01*5*10-8/(1.6*10-19)=3.12075*1012
So, we have around 3.12075*1012 electrons hit radiator per second.